Tính:
a) (27,28–4,53) : 7
b) 42,15 + 48,16: 8
c) 173,4 : 15 + 35,2 : 11
d) 89,37 –(9,6 : 6 + 15,72)
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Lời giải:
Đặt $\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$
Ta có:
\(\frac{7a-11c}{7b-11d}=\frac{7bt-11dt}{7b-11d}=\frac{t(7b-11d)}{7b-11d}=t(1)\)
\(\frac{7a+11c}{7b+11d}=\frac{7bt+11dt}{7b+11d}=\frac{t(7b+11d)}{7b+11d}=t(2)\)
Từ $(1);(2)\Rightarrow \frac{7a-11c}{7b-11d}=\frac{7a+11c}{7b+11d}$
\(a,\dfrac{7}{15}+\dfrac{4}{5}=\dfrac{7+4\times3}{15}=\dfrac{19}{15}\\ b,\dfrac{2}{3}-\dfrac{3}{8}=\dfrac{2\times8-3\times3}{24}=\dfrac{7}{24}\\ c,\dfrac{4}{5}\times\dfrac{5}{8}=\dfrac{1\times1}{2\times1}=\dfrac{1}{2}\\ d,\dfrac{1}{2}:\dfrac{2}{5}=\dfrac{1}{2}\times\dfrac{5}{2}=\dfrac{5}{4}\)
7 / 4 - 5/8 = 14 / 8 - 5 / 8 = 9/8
9 / 1 . 3 / 11 = 27 / 11
1/2 . 5 / 2 = 5 / 4
\(\dfrac{7}{4}-\dfrac{5}{8}=\dfrac{14}{8}-\dfrac{5}{8}=\dfrac{9}{8}\)
\(9\times\dfrac{3}{11}=\dfrac{9\times3}{11}=\dfrac{27}{11}\)
\(\dfrac{1}{2}:\dfrac{2}{5}=\dfrac{1}{2}\times\dfrac{5}{2}=\dfrac{5}{4}\)
\(a.\left(27,28-4,53\right):7\) \(b.42,15+48,16:8\)
\(=22,75:7=3,25\) \(=42,15+\left(48,16:8\right)\)
\(=42,15+6,02=48,17\)
\(c.173,4:15+35,2:11\) \(d.89,37-\left(9,6:6+15,72\right)\)
\(=\left(173,4:15\right)+\left(35,2:11\right)\) \(=89,37-\left(1,6+15,72\right)\)
\(=11,56+3,2=14,76\) \(=89,37-17,32=72,05\)